Differentiate $y= x(4-9x^4)^4$ So the textbook tells me to do the following question but I can't seem to get the right answer. (answer: $(4-9x^4)^3(4-45x^4)$
This is what I've done: 
apply the product rule
$u = x$
$u' = 1$
$v = (4-9x^4)^4$
$v' = -144x^3 (4-9x^4)^3$
I used the product rule and got: $(4-9x^4)^3 (-153x^4+ 4)$
What am I doing wrong? 
edit: the textbook has the wrong answer, thanks anyways :D 
 A: The textbook is wrong. I checked your answer myself and also let Wolfram Alpha do the same. Your answer is right.
Also we do not multiply $uv'$ with $vu'$ in product rule. We add them though I noticed you did not make this mistake.
A: $y = x\left(4-9x^4\right)^4$
Use the product rule, $\left(uv\right)' = u'v + uv'$, setting:
$u = x$, $v = \left(4-9x^4\right)^4$.
Then,
$u' = 1$
Using the chain rule on $v$, we get:
$v' = 4\left(4-9x^4\right)^3 \cdot \left(-36x^3\right) = -144x^3\left(4-9x^4\right)^3$
Now, $\left(uv\right)' = u'v + uv' = \left(4-9x^4\right)^4 + -144x^4\left(4-9x^4\right)^3 = \left(4-9x^4\right)^3\left(4-9x^4-144x^4\right)$.
The answer is therefore $\left(4-9x^4\right)^3\left(4-153x^4\right)$, so you were correct, assuming the problem was transcribed correctly.
A: Notice, we have $$y=x(4-9x^4)^4$$
$$\frac{dy}{dx}y=\frac{d}{dx}(x(4-9x^4)^4)$$
 applying product rule we get $$\frac{dy}{dx}=4x(4-9x^4)^3(-9\cdot 4x^3)+(4-9x^4)^4(1)$$
$$\frac{dy}{dx}=(4-9x^4)^3(-144x^4+4-9x^4)$$ $$ =(4-153x^4)(4-9x^4)^3$$
